Question 337592
You can solve for the approximation using the general equation of the parabola,
{{{y=ax^2+bx+c}}} and three points from the parabola.
Generate 3 equations with the three points (18,4),(24,10)(12,2) 
(18,4)
{{{4=a(18)^2+18b+c}}}
1.{{{324a+18b+c=4}}}
.
.
.
(24,10)
{{{10=a(24)^2+b(24)+c}}}
2.{{{576a+24b+c=10}}}
.
.
.
(12,2)
{{{2=a(12)^2+b(12)+c}}}
3.{{{144a+12b+c=2}}}
Solve for {{{a}}},{{{b}}} and {{{c}}}.
I used matrix inversion.
{{{a=1/18}}}
{{{b=-4/3}}}
{{{c=10}}}
.
.
.
{{{highlight(y=(1/18)x^2-(4/3)x+10)}}}
.
.
.
{{{drawing(300,300,-4,26,-2,12,grid(1),circle(12,2,.5),circle(18,4,0.5),circle(24,10,0.5),graph(300,300,-4,26,-2,12,
(1/18)x^2-(4/3)x+10))}}}